Calculate radial acceleration at top of chimney This question is from HRW's Fundamentals of Physics. It goes:

A tall, cylindrical chimney falls over when its base
  is ruptured. Treat the chimney as a thin rod of length $55.0\;m$. At the
  instant it makes an angle of $35.0^\circ$ with the vertical as it falls, what
  is the radial acceleration of the top?

What I did is this:
The energy of the top-most point at the chimney must've been conserved. Therefore:
$$K_0 + U_0 = K + U$$
$$\Rightarrow mgl = \frac{1}{2}mv^2 + mgl\cos{35^\circ}$$
$$\Rightarrow \frac{v^2}{l} = 2g(1 - \cos{35^\circ})$$
Now $\frac{v^2}{l}$ is the radial acceleration, considering $l$ to be the length of the chimney, in other words the radius of the circle which is the path that the top of the chimney takes. Calculating that expression gives us $\frac{v^2}{l} = a_r = 3.54\;m/s^2$. However the correct answer according to the book is $a_r = 5.32\;m/s^2$.
Where did I make a wrong assumption?
 A: You made the wrong assumption when you said:

The energy of the top-most point at the chimney must've been conserved.

Energy is only conserved for isolated systems. Your assumption would be correct if the top of the chimney was disconnected from the remainder. However, there are forces within the chimney that distribute energy from one part to the other.
You will need to use the moment of inertia of the solid rod (therefore the hint in the problem) and use angular momentum and torque.
A: For completeness, I shall post the actual solution. The energy of the entire chimney is conserved, therefore:
$$K_{0_{chimney}} + U_{0_{chimney}} = K_{chimney} + U_{chimney}$$
$$\Rightarrow mg\frac{l}{2} = \frac{1}{2}I\omega^2 + mg\frac{l}{2}\cos35^\circ$$
$$\Rightarrow mgl = \frac{1}{3}ml^2\omega^2 + mgl\cos35^\circ$$
$$\Rightarrow \frac{1}{3}l\omega^2 = g(1 - \cos35^\circ)$$
$$l\omega^2 = 3g(1 - \cos35^\circ)$$
The angular velocity $\omega$ is the same at all points of the chimney, including at the top, and therefore we have arrived at the solution, for the radial acceleration at the top is given by $a_r = l\omega^2$. So the value of the acceleration is $l\omega^2 = 3g(1 - \cos35^\circ)$ which indeed evaluates to $5.32\;m/s^2$.
